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    • CommentRowNumber1.
    • CommentAuthorAdrien Brochier
    • CommentTimeOct 16th 2021
    • (edited Oct 16th 2021)

    Let p:XYp:X\rightarrow Y be a covering and f:XZf:X \rightarrow Z be a map. What is the correct definition of the space of factorizations of ff through YY up to homotopy ? I’d like a definition not as a space of maps YZY \rightarrow Z that makes the diagram homotopy commutative, but rather something like a spaces of homotopies from ff to a map that factors on the nose.. It’s probably super elementary but for some reasons it gets me confused.

    Basically this is the topological picture generalizing: given an inclusion of groups HGH \subset G and a morphism HKH \rightarrow K, classify the lifts GKG\rightarrow K along this map (either on the nose, or up to conjugation, depending of whether we’re taking (un)pointed spaces and maps).

    Now suppose that the normal closure of HH in GG is the whole of GG. Say, to simplify, there is another subgroup HH' of GG which is conjugated to HH and such that H,HH,H' generate GG. Now assume I also have a morphism HKH' \rightarrow K (modulo details this is typically the sort of things I’d get from the topological picture, like if I have a morphism into KK out of the fundamental groupoid of XX with respect to the fiber over a chosen basepoint in YY, e.g. if i know alll those points are sent to the same point in ZZ). Now the question of whether these morphisms extends to a morphism from GG becomes a condition, not extra structure (ie we have a surjective map from the free product H*HH*H' to GG and a map from H*HH*H' to KK so now the question is whether it factors through a quotient (rather than extends along an inclusion)). In particular if it does factors through it does so in a unique way.

    What is the correct way to model this topologically, or groupoidally ? In what setting can I use that the normal closure of HH is GG to get some sort of unicity of factorization of some map ? I guess what I’m really asking is whether thee is a name for a functor F:CDF:C\rightarrow D such that End(d)End(d) is generated by the images of the End(c)End(c) for all cc s.t. F(c)=dF(c)=d or something like that…

    • CommentRowNumber2.
    • CommentAuthorUrs
    • CommentTimeOct 16th 2021

    Focusing on just the first paragraph (as I’d need to think more on what you say after “to simplify” :-)

    The situation is this extension problem:

    X f Z cov ? Y \array{ X & \xrightarrow{f} & Z \\ {}^{\mathllap{cov}} \big\downarrow & \nearrow_{ \mathrlap{\exists ?} } \\ Y }

    I take it that the left map being a “covering” map is meant to say that it’s a principal bundle for some discrete group SS? (Let me write SS for “structure group”.)

    In that case, the vertical map is the quotient by the action of SS. In the 1-category of topological spaces this means that the extension exists and then uiquely, if and only if ff is equivariant, i.e. here: invariant, since the corresponding SS action on ZZ is taken to be trivial.

    In the \infty-category of spaces the analogous statement holds in the homotopy coherent sense.

    What you are after, I gather, is a hands-on 1-category-theoretic model for this homotopy coherent situation.

    The model category for this purpose is the Borel model structure (which works with spaces modeled as simplicial sets).

    Using the Quillen equivalence here says that the derived mapping space space of homotopically SS-equivariant maps ff is equivalently the derived mapping space of the slice sSet /W¯SsSet_{/\overline{W}S} from YW¯SY \to \overline{W}S to Z×W¯W¯Z \times \overline{W} \to \overline{W} (here W¯S\overline{W}S is the simplicial classifying space and YW¯SY \to \overline{W}S is the map that classifies the covering), and this is equivalently the derived mapping space in sSetsSet from YY to ZZ.

    Now the SS-action on XX is free, by assumption of it being a covering space, which by this Prop means that XX is cofibrant in the Borel model structure. Therefore the derived equivariant mapping space out of XX is modeled by the mapping complex of naively SS-invariant maps.

    In conclusion, up to homotopy the maps ff that do extend to YY are those which have a naively SS-invariant representative, and these then extend uniquely.

    I think.

    • CommentRowNumber3.
    • CommentAuthorGuest
    • CommentTimeOct 16th 2021

    Hi Urs, thanks a lot for your comment. The thing is it’s precisely not a principal bundle for some discrete group, because in my case it’s not a regular covering. In other words, π 1(X)\pi_1(X) is not normal in π 1(Y)\pi_1(Y) (and indeed the particular situation I have a hard time NPOV-ing is the situation where its normal closure is the whole of π 1(Y)\pi_1(Y)). That being said, even in the situation you describe there’s a small subtlety (I’m cutting hairs but it got me confused for a while). Namely it’s not quite true that ff factors uniquely, in the sense that I can have two maps s,t:YZs,t:Y\rightarrow Z which are not homotopic, but such that their precomposition with pp are both homotopic to ff. In other words, a fixed ff can be homotopic to several points in the set of SS-invariants functions. Of course what you’re saying is still correct, but basically the set of solutions of the problem is, to me, the subspace of SS-invariants functions which are homotopic to ff, I guess modulo SS-equivariant homotopies. So in this sense the solution is not in general unique, which again is to be expected since it’s essentially equivalent to the pb of extending the morphism π 1(X)\pi_1(X)\rightarrow \pi_1(Z)inducedby induced by ftoamorphism to a morphism \pi_1(Y) \rightarrow \pi_1(Z)alongtheinclusioninducedby along the inclusion induced by p$, and this problem doesn’t have a unique solution in general.

    In case the covering is not regular, I also agree this should have something to do with homotopy fixed points for the π 1(Y)\pi_1(Y) action on the space fo maps XZX\rightarrow Z….

    • CommentRowNumber4.
    • CommentAuthorAdrien Brochier
    • CommentTimeOct 16th 2021
    • (edited Oct 16th 2021)

    [Sorry got logged out while I was typing… now i can’t edit, feel free to erase the previous message]

    Hi Urs,

    thanks a lot for your comment. The thing is it’s precisely not a principal bundle for some discrete group, because in my case it’s not a regular covering. In other words, π 1(X)\pi_1(X) is not normal in π 1(Y)\pi_1(Y) (and indeed the particular situation I have a hard time NPOV-ing is the situation where its normal closure is the whole of π 1(Y)\pi_1(Y)). That being said, even in the situation you describe there’s a small subtlety (I’m cutting hairs but it got me confused for a while). Namely it’s not quite true that ff factors uniquely, in the sense that I can have two maps s,t:YZs,t:Y\rightarrow Z which are not homotopic, but such that their precomposition with pp are both homotopic to ff. In other words, a fixed ff can be homotopic to several points in the space of SS-invariants functions, which are however not themselves SS-equivariantly homotopic.

    Of course what you’re saying is still correct, but basically the set of solutions of the problem is, to me, the subspace of SS-invariants functions which are homotopic to ff, I guess modulo SS-equivariant homotopies. So in this sense the solution is not in general unique, which again is to be expected since it’s essentially equivalent to the pb of extending the morphism π 1(X)π 1(Z)\pi_1(X)\rightarrow \pi_1(Z) induced by ff to a morphism π 1(Y)π 1(Z)\pi_1(Y) \rightarrow \pi_1(Z) along the inclusion induced by pp, and this problem doesn’t have a unique solution in general. i guess all i’m saying is that although the map Map(Y,Z)=Map S(X,Z)Map(X,Z)Map(Y,Z)=Map_S(X,Z)\rightarrow Map(X,Z) is injective exactly for the reasons you explained, the induced map between their π 0\pi_0 is not.

    In case the covering is not regular, I also agree this should have something to do with homotopy fixed points for the π 1(Y)\pi_1(Y) action on the space fo maps XZX\rightarrow Z….

  1. I think I’ve isolated the claim doing what I want, does anyone know a reference for this ? Let ρ:GH\rho:G\rightarrow H be a morphism of groups, and assume that HH is normally generated by imρ\im \rho. Then the induced map */G*/H*/G \rightarrow */H is in fact an epimorphism in the (2,1)-category of groupoids. In particular I believe a covering map XYX \rightarrow Y such that π 1(X)\pi_1(X) normally generates π 1(Y)\pi_1(Y) should be an epimorphism in the (,1)(\infty,1) category of spaces..

    • CommentRowNumber6.
    • CommentAuthorUrs
    • CommentTimeOct 18th 2021

    I see. Sorry, I don’t know.

    I never thought much about \infty-epimorphisms. But you are now essentially asking for the simplest non-trivial examples of these, and one would hope that somebody has thought about that.

    Looking at the couple of references we have in that entry, Raptis 2017 does connect \infty-epimorphisms to conditions on normal subgroups, e.g in Lemma 3.4.

    (This just by scanning for keywords, I don’t know if this is relevant for your question.)

  2. Indeed, I've seen that reference and will look at it, thanks ! That being said my previous claim does not seem to be true, which is confusing...
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